The allowable-stress method requires that all the uncertainties associated with the design of a structure or machine element be grouped into a single factor of safety. An alternative method of design makes it possible to distinguish between the uncertainties associated with the structure itself and those associated with the load it is designed to support. Called Load and Resistance Factor Design (LRFD), this method allows the designer to distinguish between uncertainties associated with the live load, PL (i.e., the active or time-varying load to be supported by the struc- ture) and the dead load, PD (i.e., the self weight of the structure contribut- ing to the total load).
Using the LRFD method the ultimate load, PU, of the structure (i.e., the load at which the structure ceases to be useful) should be deter- mined. The proposed design is acceptable if the following inequality is satisfied:
gDPD1gLPL#fPU (1.27) The coefficient f is the resistance factor, which accounts for the uncertain- ties associated with the structure itself and will normally be less than 1.
The coefficients gD and gL are the load factors; they account for the uncertainties associated with the dead and live load and normally will be greater than 1, with gL generally larger than gD. While a few examples and assigned problems using LRFD are included in this chapter and in Chaps. 5 and 10, the allowable-stress method of design is primarily used in this text.
Sample Problem 1.3
Two loads are applied to the bracket BCD as shown. (a) Knowing that the control rod AB is to be made of a steel having an ultimate normal stress of 600 MPa, determine the diameter of the rod for which the factor of safety with respect to failure will be 3.3. (b) The pin at C is to be made of a steel having an ultimate shearing stress of 350 MPa.
Determine the diameter of the pin C for which the factor of safety with respect to shear will also be 3.3. (c) Determine the required thickness of the bracket supports at C, knowing that the allowable bearing stress of the steel used is 300 MPa.
STRATEGY: Consider the free body of the bracket to determine the force P and the reaction at C. The resulting forces are then used with the allowable stresses, determined from the factor of safety, to obtain the required dimensions.
MODELING: Draw the free-body diagram of the hanger (Fig. 1), and the pin at C (Fig. 2).
ANALYSIS:
Free Body: Entire Bracket. Using Fig. 1, the reaction at C is represented by its com ponents Cx and Cy.
1 loMC 5 0: P(0.6 m) 2 (50 kN)(0.3 m) 2 (15 kN)(0.6 m) 5 0 P 5 40 kN
oFx 5 0: Cx 5 40 kN
oFy 5 0: Cy 5 65 kN C52C2x1C2y576.3 kN a. Control Rod AB. Since the factor of safety is 3.3, the allowable stress is
sall5 sU
F.S.5600 MPa
3.3 5181.8 MPa For P 5 40 kN, the cross-sectional area required is
Areq5 P
sall5 40 kN
181.8 MPa522031026 m2 Areq5p
4 dAB2 522031026 m2 dab 5 16.74 mm ◀
b. Shear in Pin C. For a factor of safety of 3.3, we have
tall5 tU
F.S.5350 MPa
3.3 5106.1 MPa
(continued)
t t
A
D B
dAB
C 0.6 m
0.3 m 0.3 m 50 kN 15 kN P
50 kN 15 kN 0.6 m
0.3 m 0.3 m D B
C P
Cx Cy
C
C dC
F2 F1 F1 F2 12
Fig. 1 Free-body diagram of bracket.
Fig. 2 Free-body diagram of pin at point C.
36 Introduction—Concept of Stress
As shown in Fig. 2 the pin is in double shear. We write
Areq5Cy2
tall 5 176.3 kN2y2
106.1 MPa 5360 mm2 Areq5p
4 dC25360 mm2 dC521.4 mm Use: dC 5 22 mm ◀ c. Bearing at C. Using d 5 22 mm, the nominal bearing area of each bracket is 22t. From Fig. 3 the force carried by each bracket is C/2 and the allowable bearing stress is 300 MPa. We write
Areq5Cy2
sall 5176.3 kN2y2
300 MPa 5127.2 mm2
Thus, 22t 5 127.2 t 5 5.78 mm Use: t 5 6 mm ◀ REFLECT and THINK: It was appropriate to design the pin C first and then its bracket, as the pin design was geometrically dependent upon diameter only, while the bracket design involved both the pin diameter and bracket thickness.
Sample Problem 1.4
The rigid beam BCD is attached by bolts to a control rod at B, to a hydraulic cylinder at C, and to a fixed support at D. The diameters of the bolts used are: dB 5 dD 5 38 in., dC 5 12 in. Each bolt acts in double shear and is made from a steel for which the ultimate shearing stress is tU 5 40 ksi. The control rod AB has a diameter dA5 167 in. and is made of a steel for which the ultimate tensile stress is sU 5 60 ksi. If the minimum factor of safety is to be 3.0 for the entire unit, determine the largest upward force that may be applied by the hydraulic cylinder at C.
STRATEGY: The factor of safety with respect to failure must be 3.0 or more in each of the three bolts and in the control rod. These four independent criteria need to be considered separately.
MODELING: Draw the free-body diagram of the bar (Fig. 1) and the bolts at B and C (Figs. 2 and 3). Determine the allowable value of the force C based on the required design criteria for each part.
ANALYSIS:
Free Body: Beam BCD. Using Fig. 1, first determine the force at C in terms of the force at B and in terms of the force at D.
1loMD 5 0: B114 in.22C18 in.2 50 C51.750B (1) 1loMB5 0: 2D114 in.21C16 in.2 50 C52.33D (2) (continued)
C D
B A
6 in.
8 in.
D D B
C
B C
6 in. 8 in.
Fig. 1 Free-body diagram of beam BCD.
Fig. 3 Bearing loads at bracket support at point C.
d 22 mm
t 12C
1C
2
Control Rod. For a factor of safety of 3.0
sall5 sU
F.S.560 ksi
3.0 520 ksi Th e allowable force in the control rod is
B5sall1A25120 ksi214p1167 in.2253.01 kips Using Eq. (1), the largest permitted value of C is
C51.750B51.75013.01 kips2 C 5 5.27 kips ◀
Bolt at B. tall5tUyF.S. 5 (40 ksi)y3 5 13.33 ksi. Since the bolt is in double shear (Fig. 2), the allowable magnitude of the force B exerted on the bolt is
B52F1521tall A2 52113.33 ksi2114p2138 in.2252.94 kips From Eq. (1), C51.750B51.75012.94 kips2 C 5 5.15 kips ◀
Bolt at D. Since this bolt is the same as bolt B, the allowable force is D 5 B 5 2.94 kips. From Eq. (2)
C52.33D52.3312.94 kips2 C 5 6.85 kips ◀ Bolt at C. We again have tall5 13.33 ksi. Using Fig. 3, we write
C52F2521tall A2 52113.33 ksi2114p2112 in.22 C 5 5.23 kips ◀
C
F2 F2
1 2in.
C ⫽ 2F2
Fig. 3 Free-body diagram of pin at point C.
F1
F1
B
3 8in.
B 2F1
Fig. 2 Free-body diagram of pin at point B.
Summary. We have found separately four maximum allowable val- ues of the force C. In order to satisfy all these criteria, choose the
smallest value. C 5 5.15 kips ◀
REFLECT and THINK: This example illustrates that all parts must satisfy the appropriate design criteria, and as a result, some parts have more capacity than needed.
Problems
1.29 Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that P 5 11 kN, determine the normal and shearing stresses in the glued splice.
1.30 Two wooden members of uniform rectangular cross section are joined by the simple glued scarf splice shown. Knowing that the maximum allowable shearing stress in the glued splice is 620 kPa, determine (a) the largest load P that can be safely applied, (b) the corresponding tensile stress in the splice.
1.31 The 1.4-kip load P is supported by two wooden members of uni- form cross section that are joined by the simple glued scarf splice shown. Determine the normal and shearing stresses in the glued splice.
1.32 Two wooden members of uniform cross section are joined by the simple scarf splice shown. Knowing that the maximum allowable tensile stress in the glued splice is 75 psi, determine (a) the larg- est load P that can be safely supported, (b) the corresponding shearing stress in the splice.
1.33 A centric load P is applied to the granite block shown. Knowing that the resulting maximum value of the shearing stress in the block is 2.5 ksi, determine (a) the magnitude of P, (b) the orienta- tion of the surface on which the maximum shearing stress occurs, (c) the normal stress exerted on that surface, (d) the maximum value of the normal stress in the block.
Fig. P1.29 and P1.30 75 mm
150 mm 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45 45
P'
P
Fig. P1.31 and P1.32 60
5.0 in.
3.0 in.
P' P
Fig. P1.33 and P1.34 6 in.
6 in.
P
1.34 A 240-kip load P is applied to the granite block shown. Deter- mine the resulting maximum value of (a) the normal stress, (b) the shearing stress. Specify the orientation of the plane on
thick plate by welding along a helix that forms an angle of 208 with a plane perpendicular to the axis of the pipe. Knowing that a 300-kN axial force P is applied to the pipe, determine the nor- mal and shearing stresses in directions respectively normal and tangential to the weld.
1.36 A steel pipe of 400-mm outer diameter is fabricated from 10-mm- thick plate by welding along a helix that forms an angle of 208 with a plane perpendicular to the axis of the pipe. Knowing that the maximum allowable normal and shearing stresses in the directions respectively normal and tangential to the weld are s 5 60 MPa and t 5 36 MPa, determine the magnitude P of the largest axial force that can be applied to the pipe.
1.37 A steel loop ABCD of length 5 ft and of 38-in. diameter is placed as shown around a 1-in.-diameter aluminum rod AC. Cables BE and DF, each of 12-in. diameter, are used to apply the load Q.
Knowing that the ultimate strength of the steel used for the loop and the cables is 70 ksi, and that the ultimate strength of the alu- minum used for the rod is 38 ksi, determine the largest load Q that can be applied if an overall factor of safety of 3 is desired.
Fig. P1.35 and P1.36 208 Weld
10 mm
Fig. P1.37 12 in.
9 in. 1 in.
C
D Q
A 9 in.
12 in.
F Q' B E
1in.
2 3in.
8
1.38 Link BC is 6 mm thick, has a width w 5 25 mm, and is made of a steel with a 480-MPa ultimate strength in tension. What is the factor of safety used if the structure shown was designed to sup- port a 16-kN load P?
1.39 Link BC is 6 mm thick and is made of a steel with a 450-MPa ultimate strength in tension. What should be its width w if the structure shown is being designed to support a 20-kN load P with
a factor of safety of 3? Fig. P1.38 and P1.39
A B
C
D 480 mm
908 w
P
1.40 Members AB and BC of the truss shown are made of the same alloy. It is known that a 20-mm-square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was re - corded. If a factor of safety of 3.2 is to be achieved for both bars, determine the required cross-sectional area of (a) bar AB, (b) bar AC.
1.41 Members AB and BC of the truss shown are made of the same alloy. It is known that a 20-mm-square bar of the same alloy was tested to failure and that an ultimate load of 120 kN was recorded.
If bar AB has a cross-sectional area of 225 mm2, determine (a) the factor of safety for bar AB, (b) the cross-sectional area of bar AC if it is to have the same factor of safety as bar AB.
1.42 Link AB is to be made of a steel for which the ultimate normal stress is 65 ksi. Determine the cross-sectional area of AB for which the factor of safety will be 3.20. Assume that the link will be adequately reinforced around the pins at A and B.
Fig. P1.40 and P1.41 1.4 m
0.75 m
0.4 m B A
C
Fig. P1.42
1.4 ft 35⬚
B A
C D E
1.4 ft 1.4 ft 600 lb/ft
5 kips
1.43 Two wooden members are joined by plywood splice plates that are fully glued on the contact surfaces. Knowing that the clear- ance between the ends of the members is 6 mm and that the ultimate shearing stress in the glued joint is 2.5 MPa, determine the length L for which the factor of safety is 2.75 for the loading shown.
Fig. P1.43 16 kN
L
125 mm 6 mm
16 kN
1.44 For the joint and loading of Prob. 1.43, determine the factor of
4
plate shown to a wooden beam. Knowing that the plate will sup- port a load P 5 24 kips and that the ultimate shearing stress for the steel used is 52 ksi, determine the factor of safety for this design.
Fig. P1.45 and P1.46 P
Fig. P1.47
1 2
40 mm
d
c b P
1 2P
1.46 Three steel bolts are to be used to attach the steel plate shown to a wooden beam. Knowing that the plate will support a load P 5 28 kips, that the ultimate shearing stress for the steel used is 52 ksi, and that a factor of safety of 3.25 is desired, determine the required diameter of the bolts.
1.47 A load P is supported as shown by a steel pin that has been inserted in a short wooden member hanging from the ceiling.
The ultimate strength of the wood used is 60 MPa in tension and 7.5 MPa in shear, while the ultimate strength of the steel is 145 MPa in shear. Knowing that b 5 40 mm, c 5 55 mm, and d 5 12 mm, determine the load P if an overall factor of safety of 3.2 is desired.
1.48 For the support of Prob. 1.47, knowing that the diameter of the pin is d 5 16 mm and that the magnitude of the load is P 5 20 kN, determine (a) the factor of safety for the pin (b) the required values of b and c if the factor of safety for the wooden member is the same as that found in part a for the pin.
1.49 A steel plate 14 in. thick is embedded in a concrete wall to anchor a high-strength cable as shown. The diameter of the hole in the plate is 34 in., the ultimate strength of the steel used is 36 ksi. and the ultimate bonding stress between plate and concrete is 300 psi. Knowing that a factor of safety of 3.60 is desired when P 5 2.5 kips, determine (a) the required width a of the plate, (b) the minimum depth b to which a plate of that width should be embedded in the concrete slab. (Neglect the normal stresses between the concrete and the end of the plate.)
Fig. P1.49 a
b
P
3 4in.
1 4in.
1.50 Determine the factor of safety for the cable anchor in Prob. 1.49 when P 5 2.5 kips, knowing that a 5 2 in. and b 5 6 in.
1.51 Link AC is made of a steel with a 65-ksi ultimate normal stress and has a 14312-in. uniform rectangular cross section. It is connected to a support at A and to member BCD at C by 34-in.-diameter pins, while member BCD is connected to its support at B by a
5
16-in.-diameter pin. All of the pins are made of a steel with a 25-ksi ultimate shearing stress and are in single shear. Knowing that a factor of safety of 3.25 is desired, determine the largest load P that can be applied at D. Note that link AC is not reinforced around the pin holes.
1.52 Solve Prob. 1.51, assuming that the structure has been redesigned to use 165-in.-diameter pins at A and C as well as at B and that no other changes have been made.
1.53 Each of the two vertical links CF connecting the two horizontal members AD and EG has a 10 3 40-mm uniform rectangular cross section and is made of a steel with an ultimate strength in tension of 400 MPa, while each of the pins at C and F has a 20-mm diameter and are made of a steel with an ultimate strength in shear of 150 MPa. Determine the overall factor of safety for the links CF and the pins connecting them to the horizontal members.
1.54 Solve Prob. 1.53, assuming that the pins at C and F have been Fig. P1.51
P 6 in.
8 in.
4 in.
1 2in.
A
B C D
24 kN C
A B
E
D
F G
250 mm
400 mm
250 mm
12-mm-diameter pins are used at B and D. Knowing that the ulti- mate shearing stress is 100 MPa at all connections and that the ultimate normal stress is 250 MPa in each of the two links joining B and D, determine the allowable load P if an overall factor of safety of 3.0 is desired.
Fig. P1.55
180 mm 200 mm
Top view
Side view Front view
8 mm
20 mm
8 mm 8 mm
12 mm 12 mm
B C
B
D D
A
B C
A
P
1.56 In an alternative design for the structure of Prob. 1.55, a pin of 10-mm-diameter is to be used at A. Assuming that all other speci- fications remain unchanged, determine the allowable load P if an overall factor of safety of 3.0 is desired.
*1.57 A 40-kg platform is attached to the end B of a 50-kg wooden beam AB, which is supported as shown by a pin at A and by a slender steel rod BC with a 12-kN ultimate load. (a) Using the Load and Resistance Factor Design method with a resistance factor f 5 0.90 and load factors gD 5 1.25 and gL 5 1.6, deter- mine the largest load that can be safely placed on the platform.
(b) What is the corresponding conventional factor of safety for rod BC ?
*1.58 The Load and Resistance Factor Design method is to be used to select the two cables that will raise and lower a platform support- ing two window washers. The platform weighs 160 lb and each of the window washers is assumed to weigh 195 lb with equip- ment. Since these workers are free to move on the platform, 75%
of their total weight and the weight of their equipment will be used as the design live load of each cable. (a) Assuming a resis- tance factor f 5 0.85 and load factors gD5 1.2 and gL 5 1.5, determine the required minimum ultimate load of one cable.
(b) What is the corresponding conventional factor of safety for the selected cables?
Fig. P1.57 1.8 m
2.4 m
A B
C
Fig. P1.58
P P
Review and Summary
This chapter was devoted to the concept of stress and to an introduction to the methods used for the analysis and design of machines and load- bearing structures. Emphasis was placed on the use of a free-body diagram to obtain equilibrium equations that were solved for unknown reactions.
Free-body diagrams were also used to find the internal forces in the vari- ous members of a structure.
Axial Loading: Normal Stress
The concept of stress was first introduced by considering a two-force member under an axial loading. The normal stress in that member (Fig. 1.41) was obtained by
s5P
A (1.5)
The value of s obtained from Eq. (1.5) represents the average stress over the section rather than the stress at a specific point Q of the section.
Considering a small area DA surrounding Q and the magnitude DF of the force exerted on DA, the stress at point Q is
s5 lim
¢Ay0
¢F
¢A (1.6)
In general, the stress s at point Q in Eq. (1.6) is different from the value of the average stress given by Eq. (1.5) and is found to vary across the section. However, this variation is small in any section away from the points of application of the loads. Therefore, the distribution of the normal stresses in an axially loaded member is assumed to be uniform, except in the immediate vicinity of the points of application of the loads.
For the distribution of stresses to be uniform in a given section, the line of action of the loads P and P9 must pass through the centroid C. Such a loading is called a centric axial loading. In the case of an eccentric axial loading, the distribution of stresses is not uniform.
Transverse Forces and Shearing Stress
When equal and opposite transverse forces P and P9 of magnitude P are applied to a member AB (Fig. 1.42), shearing stresses t are created over any section located between the points of application of the two forces.
A
P' P
Fig. 1.41 Axially loaded member with cross section normal to member used to define normal stress.
A C B
P
P⬘
Fig. 1.42 Model of transverse resultant forces on
be assumed to be uniform. However, dividing the magnitude P—referred to as the shear in the section—by the cross-sectional area A, the average shearing stress is:
tave5 P
A (1.8)
Single and Double Shear
Shearing stresses are found in bolts, pins, or rivets connecting two struc- tural members or machine components. For example, the shearing stress of bolt CD (Fig. 1.43), which is in single shear, is written as
tave5P A5F
A (1.9)
C
D
A F
B E' E F'
Fig. 1.43 Diagram of a single-shear joint.
K
A B
L
E H
G J
C D
K' L' F' F
Fig. 1.44 Free-body diagram of a double-shear joint.
The shearing stresses on bolts EG and HJ (Fig. 1.44), which are both in double shear, are written as
tave5P A5Fy2
A 5 F
2A (1.10)
Bearing Stress
Bolts, pins, and rivets also create stresses in the members they connect along the bearing surface or surface of contact. Bolt CD of Fig. 1.43 creates stresses on the semicylindrical surface of plate A with which it is in contact (Fig. 1.45). Since the distribution of these stresses is quite complicated, one uses an average nominal value sb of the stress, called bearing stress.
sb5P A5 P
td (1.11)
A
C
D d
t
F P
F'
Fig. 1.45 Bearing stress from force P and the single-shear bolt associated with it.
Method of Solution
Your solution should begin with a clear and precise statement of the problem. Then draw one or several free-body diagrams that will be used